Choosing Cycles for Jordan Canonical Form

Let $$T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$$. Suppose the characteristic polynomial of $$T$$ is $$f(t) = (t-\lambda)^3$$ for the eigenvalue $$\lambda$$. Suppose also that $$\dim E_\lambda = 2$$, and that $$\beta = \{v_1,v_2\}$$ is a basis for this eigenspace.

If I wanted to create the Jordan Canonical Form Matrix for $$T$$, I would require a basis $$\alpha = \beta \cup \text{span}(v_3)$$ where $$v_3 \in K_\lambda$$, the generalized Eigenspace of $$\lambda$$.

Let $$a,b$$ be vectors such that

$$(T-\lambda I)a = v_1 \qquad (T-\lambda I)b = v_2$$

Would both $$a$$ and $$b$$ be appropriate choices for $$v_3$$?

Either is an appropriate choice assuming it exists. But at most one of $$a$$ and $$b$$ can exist, and it's possible that neither exists.

You have to choose $$v_1$$ and $$v_2$$ more carefully. You should really find $$a$$ or $$b$$ first, i.e., find a solution to the set of equations $$(T-\lambda I)^2 a = 0 \\ (T-\lambda I) a \ne 0$$ and then set $$v_1 = (T-\lambda I)a$$. Then find another $$\lambda$$-eigenvector $$v_2$$ which is independent from $$v_1$$.

Also, if the characteristic polynomial is $$(t-\lambda)^3$$, then the generalized $$\lambda$$-eigenspace is all of $$\mathbb{R}^3$$. So you need a better $$v_3$$ than you claim you do.

At most one can exist:

If both existed, then $$\{v_1,v_2,a,b\}$$ would be linearly independent: if $$c_1v_1 + c_2a + d_1v_2 + d_2b = 0,$$ with $$c_1,c_2,d_1,d_2 \in \mathbb{R}$$, then apply $$(T-\lambda I)$$ to both sides, so that $$c_2v_1 + d_2v_2 = 0,$$ which implies $$c_2=d_2=0$$ since $$\{v_1,v_2\}$$ is a linearly independent set, and then $$c_1v_1 + d_1v_2 = 0,$$ and so $$c_1=d_1=0$$ for the same reason. But $$\dim(\mathbb{R}^3) < 4$$.

It's possible that neither exists:

Consider $$T=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix},$$ which has characteristic polynomial $$(t-1)^3$$. Taking $$v_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \text{ and }\; v_2= \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix},$$ the equations $$(T-I)a = v_1$$ and $$(T-I)b = v_2$$ have no solutions, but $$\{v_1,v_2\}$$ is a basis for the $$1$$-eigenspace.

• In your example where neither exists, it is guaranteed that there will exist another set of vectors, say $w_1,w_2$ that would form a basis such that one of $a,b$ will exist? Apr 3, 2019 at 17:26
• Yes, because the eigenvalues are real (not complex) and it can be proven that any such matrix can be put into Jordan normal form over $\mathbb{R}$. Take $w_1 = (1,0,0)^T$ and $w_2 = (0,1,0)^T$. Then $b = (0,0,1)^T$ is a solution to $(T-I)b = w_2$. (Hint: I put $T$ in JNF to begin with.) Apr 3, 2019 at 17:28